Let's first look at the numerator and denominator separately.

$\displaystyle x^{2}+6x+5$: We need two numbers that multiply to 5 and add to 6. The numbers 1 and 5 work. So, $\displaystyle x^{2}+6x+5 = (x+5)(x+1)$

$\displaystyle x^{2}+10x + 25$: We need two numbers that multiply to 25 and add to 10. The numbers 5 and 5 work. So, $\displaystyle x^{2}+10x + 25 = (x+5)(x+5)$

Putting this together, $\displaystyle \frac{x^{2}+6x+5}{x^{2}+10x+25} = \frac{(x+5)(x+1)}{(x+5)(x+5)} = \frac{x+1}{x+5}$

Let's combine like terms.

$\displaystyle 2x+3y-3=2x+3y$

$\displaystyle -3=0$, so the equation has no solution.

You have to isolate $\displaystyle D$ by moving around the separate components in the problem.  The steps should go as follows:

$\displaystyle P = \frac{27D}{2d - D}$

$\displaystyle (2d - D)* P = 27D$

$\displaystyle 2dP - DP = 27D$

$\displaystyle 2dP = 27D + DP$

$\displaystyle 2dP = D*(27 + P)$

$\displaystyle \frac{2dP}{27 + P} = D$

To simplify algebraically, we combine like terms. First, we should get the expression in one long string, by removing the parentheses. So remembering the communitive property, the first group in parentheses will have no changes when we remove the parentheses. So $\displaystyle (12X - 2X- 22Y - 32XY) - (42X+62Y-52XY)$ simplifies to $\displaystyle 12X - 2X- 22Y - 32XY - (42X+62Y-52XY)$

However, note the second group in parentheses is being subtracted. So we must invert all the signs in the group to simplify properly. So the previous expression simplifies to 

$\displaystyle 12X - 2X- 22Y - 32XY - 42X-62Y+52XY$

Finally we reorder and combine like terms to get

$\displaystyle 12X - 2X- 42X- 22Y -62Y- 32XY +52XY = -32X-84Y + 20XY$

The best way to illustrate the answer to this question is to do these operations to the number 1.

First, divide by 4:

$\displaystyle 1 \div 4 = 0.250$

Now move the decimal point right three spaces:

$\displaystyle 0.250\Rightarrow 250.$

This has the effect of multiplying the number by 250.

$\displaystyle \log_{7} x^{4} = 4 \log_{7} x = 4 \cdot\frac{ \ln x }{\ln 7} =\frac{ 4\ln x }{\ln 7}$

Let the value of the first integer be $\displaystyle N$.  This means that the consecutive integers will be $\displaystyle N$, $\displaystyle N+1$, and $\displaystyle N+2$.  The sum must be 12 which means that 

$\displaystyle N+(N+1)+(N+2)=12\Rightarrow 3N+3=12\Rightarrow N=3$ 

Since $\displaystyle N=3$ the consecutive integers are 3, 4, and 5.  The middle integer is 4.

$\displaystyle y=\frac{3x}{2+x}\Rightarrow y(2+x)=3x$ 

$\displaystyle 2y+xy=3x\Rightarrow xy-3x=-2y$

$\displaystyle x(y-3)=-2y\Rightarrow x=\frac{-2y}{y-3}$

Foil

$\displaystyle (2+7x)(2-7x)$

$\displaystyle 4+14x-14x-49x^2$

$\displaystyle 4-49x^2$

$\displaystyle (x^a)^b=x^{ab}$

Therefore:

$\displaystyle (x^4)^{10}=x^{40}$